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In this video, weβll learn how to use combinations to solve counting problems.
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A combination is a rearrangement of a collection of items.
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For example, imagine we have the letters π΄, π΅, πΆ, and π·.
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We could arrange two of them as π΄π΅, π΅πΆ, πΆπ·, and so on.
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Each arrangement is an example of a combination.
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But for combinations, we say order doesnβt matter.
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And so π΄π΅ would be considered the same arrangement as π΅π΄.
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Remember, of course, that if order does matter, weβre actually thinking about permutations.
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And therefore, our formula is going to be different.
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Itβs really important that weβre careful to distinguish between these.
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Our job is now to find a way to count them.
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And to help us find a formula, weβre going to begin by considering an example.
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Olivia was asked by her teacher to choose five from the eight topics given to her.
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How many different five-topic groups could she choose?
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In this question, weβre looking at how many ways we can choose five items from a group of eight.
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Now we should see quite quickly that the order here doesnβt matter.
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For example, letβs say three of her topics are fractions, decimals, percentages.
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She could choose them in that order: fractions, decimals, percentages.
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She could alternatively say fractions first and then choose percentages and then decimals.
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There are in fact six different ways that she could choose these topics.
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But we see that choosing fractions, decimals, percentages would be exactly the same as selecting decimals, then percentages, then fractions.
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When we want to choose a number of items from a larger group and order doesnβt matter, these are called combinations.
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Now, in order to find the number of combinations, weβre going to begin by thinking about permutations.
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Now, permutations occur when order does matter.
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So if we think about our earlier example, where we ordered the three topics, there was only one combination but six different permutations.
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We might recall that πPπ is the number of ways of choosing π items from a selection of π when order does matter.
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Itβs the number of permutations.
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And we calculate this by working out π factorial divided by π minus π factorial.
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So letβs begin by working out the number of permutations of five topics from a total of eight.
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Thatβs eight P five.
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Thatβs eight factorial over eight minus three factorial, which is 6720.
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There are 6720 permutations.
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But we know that order doesnβt matter.
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And so we need to find a way to get rid of the extra permutations.
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If we go back to our earlier set of three topics, we saw that that was six P three.
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There were six permutations.
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To get rid of the extra permutations, we would divide by three factorial.
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Three factorial is six, so we get six divided by six, which is equal to one.
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In the case where we choose five topics from eight, we need to divide by five factorial.
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And so the calculation that we can perform to find the number of combinations is eight factorial divided by five factorial times eight minus three factorial.
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And thatβs equal to 56.
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This gives rise to a formula.
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When finding combinations, we notice that we have less than the number of permutations.
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To take into account the number of extra times weβve counted the same set of our items, we need to divide the number of permutations by π factorial.
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The number of combinations of size π taken from a collection of π items is given by πCπ.
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And thatβs πPπ divided by π factorial, which can be written as π factorial divided by π factorial times π minus π factorial.
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πCπ is sometimes read as π choose π, and itβs also referred to as the binomial coefficient.
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You might see it written in any of these formats.
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The one you choose will be partly personal preference and partly where youβre located in the world.
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In this video, weβll primarily use these two forms.
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Letβs now consider an example that demonstrates how we can apply this formula.
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The names of four students are each written on a piece of paper, which are then placed in a hat.
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If two names are randomly selected from the hat, determine the number of all two-student selections that are possible.
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In this question, weβre looking to choose two names from a selection of four.
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And so we should be beginning to think about combinations and permutations.
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Combinations are a way of choosing rearrangements where order doesnβt matter.
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So, for example, if we were choosing out of the letters π΄, π΅, πΆ, and π·, the arrangement π΄ and π΅ would be the same as the arrangement π΅ and π΄.
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With permutations, order does matter.
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So π΄ and π΅ would be a different arrangement to π΅ and π΄.
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Letβs think about choosing the two names from a hat.
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If we choose Ali and Ben, for example, that is exactly the same as choosing Ben and Ali.
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Order doesnβt matter.
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And so we can see that we are thinking about the number of combinations.
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The number of combinations of size π taken from a collection of π items is given by πCπ.
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And the formula is π factorial divided by π factorial times π minus π factorial.
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Weβre choosing two names from a selection of four.
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So the number of combinations is four C two.
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And thatβs four factorial over two factorial times four minus two factorial.
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Now, note that we could use the πCπ button on our calculator.
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But actually, thereβs a little shortcut that can help us to calculate these by hand.
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We begin by rewriting four minus two factorial as two factorial.
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But of course, four factorial is four times three times two times one, which can further be rewritten as four times three times two factorial.
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And now we see we can divide through by a common factor of two factorial.
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In fact, two factorial is simply two, so we can divide once again by two.
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And we see four choose two is two times three divided by one, which is simply six.
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There are six ways to choose two names from the four given.
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Weβre now going to consider how we can apply this formula to a problem involving sets.
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Let π be the set containing π₯, where π₯ is an integer greater than or equal to 10 and less than or equal to 16, and π is the set containing the elements π and π, where π and π are elements of the set π₯ and π is not equal to π.
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Determine the value of π of π, where π of π is the number of elements in π.
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Letβs just begin by looking at what this set notation is actually telling us.
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Set π contains the number π₯, which is an integer greater than or equal to 10 and less than or equal to 16.
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And so π could be rewritten as the set containing the elements 10, 11, 12, 13, 14, 15, and 16.
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Then, we have set π, and set π contains all pairs of integer combinations π and π such that π is not equal to π.
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But both π and π are elements of set π.
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So itβs like a list of pairs of numbers, and we need to work out how many pairs are in that list.
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And so π of π is the number of ways of choosing two elements from a total of seven.
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And thatβs because there are seven elements in set π.
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Since π cannot be equal to π, thereβs no repetition.
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But of course, choosing 10 and 11 would be the same as choosing 11 and then 10.
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And so weβre interested in combinations.
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And π of π is seven choose two.
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The formula of π choose π, the number of ways of choosing π items, from a set of π is π factorial over π factorial times π minus π factorial.
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And so seven choose two is seven factorial divided by two factorial times seven minus two factorial.
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Now, we can write seven minus two factorial as five factorial.
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And then we can also write the numerator seven factorial as seven times six times five factorial.
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Then we divide through by five factorial.
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And since two factorial is two, we can also divide through by two.
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And seven choose two simplifies to seven times three over one, which is simply 21.
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And so π of π, which was the number of ways of choosing two unique elements from a set of seven where order doesnβt matter, is 21.
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We can also apply this formula more than once to help us answer counting problems.
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A class contains 14 boys and 13 girls.
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In how many ways can you select a team of eight people from the class such that every member of the team is of the same sex?
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Weβre told to choose a team of eight people.
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But every single member of that team must be of the same sex.
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So weβre either going to choose eight out of the 14 boys or eight out of the 13 girls.
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And so we next ask ourselves, does order matter?
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Well, no choosing, say, boy π and boy π would be the same as choosing boy π and then boy π.
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When order doesnβt matter, weβre looking at counting combinations.
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And we say that the number of ways of choosing π items from a set of π is π choose π.
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And itβs π factorial over π factorial times π minus π factorial.
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And so weβre going to begin by performing two different calculations.
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We want to work out the number of ways of choosing eight boys from a total of 14.
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So thatβs 14 choose eight.
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And then weβre separately going to work out the number of ways of choosing eight girls from the group of 13.
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Thatβs 13 choose eight.
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14 choose eight is 14 factorial over eight factorial times 14 minus eight factorial.
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And if we type that into our calculator or use the combinations button, we see that 14 choose eight is equal to 3003.
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Then 13 choose eight is 13 factorial over eight factorial times 13 minus eight factorial, which is 1287.
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What weβre trying to count is the number of ways of choosing eight boys or choosing eight girls.
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And so thatβs the sum of our individual calculations.
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Itβs 3003 plus 1287, and thatβs 4290.
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There are 4290 ways to select a team of eight people from the class so that every member of the team is of the same sex.
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In our final example, weβll consider a problem where we need to work out the value of π when given the total number of combinations and the value of π.
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At a college, there were 120 ways to select 119 students to attend a seminar.
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Determine the number of students at the college.
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Letβs begin by letting π be equal to the total number of students at the college.
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π is 119; weβre looking to select 119 students from a total of π.
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And weβre told there are 120 ways to do so.
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Now, in this case, order doesnβt matter.
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And so these are combinations.
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There are 120 combinations of students.
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The formula we use for counting combinations is πCπ equals π factorial over π factorial times π minus π factorial, where π is the number of items weβre looking to choose and π is the total number of items we can choose from.
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In this question, weβre looking to find the number of combinations of 119 students from a set of π.
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So thatβs π choose 119.
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According to our formula, thatβs π factorial over 119 factorial times π minus 119 factorial.
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But of course, we were actually told that there were 120 ways to select these students.
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So we get 120 equals π factorial over 119 factorial times π minus 119 factorial.
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Weβre now going to multiply both sides of our equation by 119 factorial.
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The left-hand side then becomes 120 times 119 factorial.
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We could also consider that to be equal to simply 120 factorial.
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And then on the right-hand side, we have π factorial over π minus 119 factorial.
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But actually, we can simplify the expression on the right-hand side somewhat.
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If we consider π factorial as being π times π minus one times π minus two all the way down to one, then weβre able to simplify this fraction by dividing the numerator and denominator by π minus 119 factorial.
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And then that leaves us with π times π minus one times π minus two all the way down to π minus 117 times π minus 118.
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We divided by π minus 119 factorial.
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So any term after this simply disappears.
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Weβre now going to use this to compare each side of our earlier equation.
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We can say that π times π minus one times π minus two all the way down to π minus 118 must be equal to this left-hand side, 120 times 119 factorial.
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Now we can write this as 120 times 119 times 118 and so on.
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Now we do need to be a little bit careful here.
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On the numerical side of our equation, we have 120 term.
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And on the algebraic side, the side with the πβs, we have 119 terms.
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But of course, this is the same as multiplying π by π minus one by π minus two and so on, all the way down to π minus 118 and then multiplying that by one.
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And so now there are two expressions equal to one another that contain the product of 120 terms.
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In each case, weβve got 119 terms that follow the same pattern, that is, a number multiplied by one less multiplied by one less and so on.
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And then each one itself is multiplied by one.
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We can now equate both sides of this equation.
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If we consider the largest terms in our equation, we could say π must equal to 120.
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Itβs not necessary to just choose this one.
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We could choose any terms in our equation.
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The next smallest term on each side is π minus one and 119.
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Equating these gives us π minus one equals 119.
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And if we add one to both sides, we get π equals 120.
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Even moving all the way down to π minus 118 and two, equating this gives us π minus 118 equals two.
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And we add 118 to both sides to give us π equals 120.
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And we can therefore say that the number of students at the college is 120.
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Weβre now going to summarize the key points from this lesson.
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In this video, weβve learned that combinations are arrangements of π items from a group of π.
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But order doesnβt matter.
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And of course, there must be no repetition.
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We can count the number of combinations, thatβs the number of selections of π items from a set of π, by working out π factorial over π factorial times π minus π factorial.
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But of course, this representation looks different depending on where in the world you come from.